The height of water, in metres, in Dungeness harbour is modelled by the function $H\left(t\right)=a \sin \left(b\right(t-c\left)\right)+d$, where $t$ is the number of hours after midnight, and $a, b, c$ and $d$ are constants, where $a>0, b>0$ and $c>0$.

The following graph shows the height of the water for $13$ hours, starting at midnight.

The first high tide occurs at $04:30$ and the next high tide occurs $12$ hours later. Throughout the day, the height of the water fluctuates between $2.2 \text{m}$ and $6.8 \text{m}$.

All heights are given correct to one decimal place.

OAB is a sector of the circle with centre O and radius $r$, as shown in the following diagram.

The angle AOB is $\mathrm{\theta <!--\; \theta \; -->}$ radians, where .

The point C lies on OA and OA is perpendicular to BC.

Consider a function $f$, such that $f\left(x\right)=5.8\text{sin}\left(\frac{\pi }{6}(x+1)\right)+b$, 0 ≤ $x$ ≤ 10, $b\in <!--\; \in \; -->\mathbb{R}$.

The function $f$ has a local maximum at the point (2, 21.8) , and a local minimum at (8, 10.2).

A second function $g$ is given by $g(x)=p\text{sin}\left(\frac{2\mathrm{\pi <!--\; \pi \; -->}}{9}\left(x-<!--\; -\; -->3.75\right)\right)+q$,  0 ≤ $x$ ≤ 10;  $p$, $q\in <!--\; \in \; -->\mathbb{R}$.

The function $g$ passes through the points (3, 2.5) and (6, 15.1).

A water container is made in the shape of a cylinder with internal height $h$ cm and internal base radius $r$ cm.

The water container has no top. The inner surfaces of the container are to be coated with a water-resistant material.

The volume of the water container is $0.5{\text{m}}^3$.

The water container is designed so that the area to be coated is minimized.

One can of water-resistant material coats a surface area of $2000\text{ c}{\text{m}}^2$.

The following diagram shows the graph of $f(x)=a\sin <!--\; \; -->bx+c$, for $0⩽<!--\; ⩽\; -->x⩽<!--\; ⩽\; -->12$.

The graph of $f$ has a minimum point at $(3,5)$ and a maximum point at $(9,17)$.

The graph of $g$ is obtained from the graph of $f$ by a translation of $\left(\begin{array}{c}k\\ 0\end{array}\right)$. The maximum point on the graph of $g$ has coordinates $(11.5,17)$.

The graph of $g$ changes from concave-up to concave-down when $x=w$.

Let $\stackrel{\to <!--\; \to \; -->}{\text{AB}}=\left(\begin{array}{c}4\\ 1\\ 2\end{array}\right)$.

Olivia’s house consists of four vertical walls and a sloping roof made from two rectangles. The height, $\text{CD}$, from the ground to the base of the roof is 4.5 m.

The base angles of the roof are $\text{A}\stackrel{\wedge <!--\; \wedge \; -->}{\text{B}}<!--\; \; -->\text{C}={27}^{\circ <!--\; \circ \; -->}$ and $\text{A}\stackrel{\wedge <!--\; \wedge \; -->}{\text{C}}<!--\; \; -->\text{B}={26}^{\circ <!--\; \circ \; -->}$.

The house is 10 m long and 5 m wide.

The length $\text{AC}$ is approximately 2.84 m.

Olivia decides to put solar panels on the roof. The solar panels are fitted to both sides of the roof.

Each panel is 1.6 m long and 0.95 m wide. All the panels must be arranged in uniform rows, with the shorter edge of each panel parallel to $\text{AB}$ or $\text{AC}$. Each panel must be at least 0.3 m from the edge of the roof and the top of the roof, $\text{AF}$.

Olivia estimates that the solar panels will cover an area of 29 m².

A company is designing a new logo. The logo is created by removing two equal segments from a rectangle, as shown in the following diagram.

The rectangle measures $5 \text{cm}$ by $4 \text{cm}$. The points $\text{A}$ and $\text{B}$ lie on a circle, with centre $\text{O}$ and radius $2 \text{cm}$, such that $\text{AÔB}=\theta$, where $0<\theta <\pi$. This information is shown in the following diagram.

Using geometry software, Pedro draws a quadrilateral $\text{ABCD}$. $\text{AB}=8 \text{cm}$ and $\text{CD}=9 \text{cm}$. Angle $\text{BAD}=51.5°$ and angle $\text{ADB}=52.5°$. This information is shown in the diagram.

$\text{CE}=7 \text{cm}$, where point $\text{E}$ is the midpoint of $\text{BD}$.

The following diagram shows a water wheel with centre $\text{O}$ and radius $10$ metres. Water flows into buckets, turning the wheel clockwise at a constant speed.

The height, $h$ metres, of the top of a bucket above the ground $t$ seconds after it passes through point $\text{A}$ is modelled by the function

$h\left(t\right)=13+8 \cos \left(\frac{\mathrm{\pi }}{18}t\right)-6 \sin \left(\frac{\mathrm{\pi }}{18}t\right)$, for $t\ge 0$.

A bucket moves around to point $\text{B}$ which is at a height of $4.06$ metres above the ground. It takes $k$ seconds for the top of this bucket to go from point $\text{A}$ to point $\text{B}$.

The chord $\left[\text{AB}\right]$ is $17.0$ metres, correct to three significant figures.

Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.

Let $\stackrel{\to <!--\; \to \; -->}{\text{PR}}$ = 6i − j + 3k.

The Tower of Pisa is well known worldwide for how it leans.

Giovanni visits the Tower and wants to investigate how much it is leaning. He draws a diagram showing a non-right triangle, ABC.

On Giovanni’s diagram the length of AB is 56 m, the length of BC is 37 m, and angle ACB is 60°. AX is the perpendicular height from A to BC.

Giovanni’s tourist guidebook says that the actual horizontal displacement of the Tower, BX, is 3.9 metres.

Consider the points A(−3, 4, 2) and B(8, −1, 5).

A line L has vector equation $r=\left(\begin{array}{c}2\\ 0\\ -<!--\; -\; -->5\end{array}\right)+t\left(\begin{array}{c}1\\ -<!--\; -\; -->2\\ 2\end{array}\right)$. The point C (5, $y$, 1) lies on line L.

Consider the lines $L_1$ and $L_2$ with respective equations

$L_1\text{:}y=-<!--\; -\; -->\frac{2}{3}x+9$  and  $L_2\text{:}y=\frac{2}{5}x-<!--\; -\; -->\frac{19}{5}$.

A third line, $L_3$, has gradient $-<!--\; -\; -->\frac{3}{4}$.

A pan, in which to cook a pizza, is in the shape of a cylinder. The pan has a diameter of 35 cm and a height of 0.5 cm.

A chef had enough pizza dough to exactly fill the pan. The dough was in the shape of a sphere.

The pizza was cooked in a hot oven. Once taken out of the oven, the pizza was placed in a dining room.

The temperature, $P$, of the pizza, in degrees Celsius, °C, can be modelled by

$$P(t)=a(2.06{)}^{-<!--\; -\; -->t}+19,t⩾<!--\; ⩾\; -->0$$

where $a$ is a constant and $t$ is the time, in minutes, since the pizza was taken out of the oven.

When the pizza was taken out of the oven its temperature was 230 °C.

The pizza can be eaten once its temperature drops to 45 °C.

The depth of water in a port is modelled by the function $d(t)=p\cos <!--\; \; -->qt+7.5$, for $0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->12$, where $t$ is the number of hours after high tide.

At high tide, the depth is 9.7 metres.

At low tide, which is 7 hours later, the depth is 5.3 metres.

The diagram shows a circle, centre O, with radius 4 cm. Points A and B lie on the circumference of the circle and AÔB = θ , where 0 ≤ θ ≤ $\mathrm{\pi <!--\; \pi \; -->}$.

A restaurant serves desserts in glasses in the shape of a cone and in the shape of a hemisphere. The diameter of a cone shaped glass is 7.2 cm and the height of the cone is 11.8 cm as shown.

The volume of a hemisphere shaped glass is $225\text{ c}{\text{m}}^3$.

The restaurant offers two types of dessert.

The regular dessert is a hemisphere shaped glass completely filled with chocolate mousse. The cost, to the restaurant, of the chocolate mousse for one regular dessert is $1.89.

The special dessert is a cone shaped glass filled with two ingredients. It is first filled with orange paste to half of its height and then with chocolate mousse for the remaining volume.

The cost, to the restaurant, of $100\text{ c}{\text{m}}^3$ of orange paste is $7.42.

A chef at the restaurant prepares 50 desserts; $x$ regular desserts and $y$ special desserts. The cost of the ingredients for the 50 desserts is $111.44.

Abdallah owns a plot of land, near the river Nile, in the form of a quadrilateral ABCD.

The lengths of the sides are $\text{AB}=\text{40 m, BC}=\text{115 m, CD}=\text{60 m, AD}=\text{84 m}$ and angle $\mathrm{B}\stackrel{^<!--\; ^\; -->}{\mathrm{A}}\mathrm{D}={90}^{\circ <!--\; \circ \; -->}$.

This information is shown on the diagram.

The formula that the ancient Egyptians used to estimate the area of a quadrilateral ABCD is

$\text{area}=\frac{(\text{AB}+\text{CD})(\text{AD}+\text{BC})}{4}$.

Abdallah uses this formula to estimate the area of his plot of land.

Two straight fences meet at point $\text{A}$ and a field lies between them.

A horse is tied to a post, $\text{P}$, by a rope of length $r$ metres. Point $\text{D}$ is on one fence and point $\text{E}$ is on the other, such that $\text{PD}=\text{PE}=\text{PA}=r$ and $\text{D}\hat{\text{P}}\text{E}=\theta$ radians. This is shown in the following diagram.

The length of the arc $\text{DE}$ shown in the diagram is $28 \text{m}$.

A new fence is to be constructed between points $\text{B}$ and $\text{C}$ which will enclose the field, as shown in the following diagram.

Point $\text{C}$ is due west of $\text{B}$ and $\text{AC}=800 \text{m}$ . The bearing of $\text{B}$ from $\text{A}$ is $195°$.

Note: In this question, distance is in millimetres.

Let $f(x)=x+a\sin <!--\; \; -->\left(x-<!--\; -\; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}\right)+a$, for $x⩾<!--\; ⩾\; -->0$.

The graph of $f$ passes through the origin. Let ${\text{P}}_k$ be any point on the graph of $f$ with $x$-coordinate $2k\mathrm{\pi <!--\; \pi \; -->}$, where $k\in <!--\; \in \; -->\mathbb{N}$. A straight line $L$ passes through all the points ${\text{P}}_k$.

Diagram 1 shows a saw. The length of the toothed edge is the distance AB.

The toothed edge of the saw can be modelled using the graph of $f$ and the line $L$. Diagram 2 represents this model.

The shaded part on the graph is called a tooth. A tooth is represented by the region enclosed by the graph of $f$ and the line $L$, between ${\text{P}}_k$ and ${\text{P}}_{k+1}$.

A farmer is placing posts at points $\text{A}$, $\text{B}$, and $\text{C}$ in the ground to mark the boundaries of a triangular piece of land on his property.

From point $\text{A}$, he walks due west $230$ metres to point $\text{B}$.
From point $\text{B}$, he walks $175$ metres on a bearing of $063°$ to reach point $\text{C}$.

This is shown in the following diagram.

The farmer wants to divide the piece of land into two sections. He will put a post at point $\text{D}$, which is between $\text{A}$ and $\text{C}$. He wants the boundary $\text{BD}$ to divide the piece of land such that the sections have equal area. This is shown in the following diagram.

Let $f\left(x\right)=12\text{cos}x-<!--\; -\; -->5\text{sin}x,-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}⩽<!--\; ⩽\; -->x⩽<!--\; ⩽\; -->2\mathrm{\pi <!--\; \pi \; -->}$, be a periodic function with $f\left(x\right)=f\left(x+2\mathrm{\pi <!--\; \pi \; -->}\right)$

The following diagram shows the graph of $f$.

There is a maximum point at A. The minimum value of $f$ is −13 .

A ball on a spring is attached to a fixed point O. The ball is then pulled down and released, so that it moves back and forth vertically.

The distance, d centimetres, of the centre of the ball from O at time t seconds, is given by

$d\left(t\right)=f\left(t\right)+17,0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->5.$

The six blades of a windmill rotate around a centre point $\text{C}$. Points $\text{A}$ and $\text{B}$ and the base of the windmill are on level ground, as shown in the following diagram.

From point $\text{A}$ the angle of elevation of point $\text{C}$ is $0.6$ radians.

An observer walks $7$ metres from point $\text{A}$ to point $\text{B}$.

The observer keeps walking until he is standing directly under point $\text{C}$. The observer has a height of $1.8$ metres, and as the blades of the windmill rotate, the end of each blade passes $2.5$ metres over his head.

One of the blades is painted a different colour than the others. The end of this blade is labelled point $\text{D}$. The height $h$, in metres, of point $\text{D}$ above the ground can be modelled by the function $h\left(t\right)=p \cos \left(\frac{3\mathrm{\pi }}{10}t\right)+q$, where $t$ is in seconds and $p, q\in \mathrm{ℝ}$. When $t=0$, point $\text{D}$ is at its maximum height.

The following diagram shows quadrilateral ABCD.

$\text{AB}=11\text{cm,}\text{BC}=6\text{cm,}\text{B}\stackrel{\wedge <!--\; \wedge \; -->}{\text{A}}<!--\; \; -->{\text{D = 100}}^{\circ <!--\; \circ \; -->}\text{, and C}\stackrel{\wedge <!--\; \wedge \; -->}{\text{B}}<!--\; \; -->{\text{D = 82}}^{\circ <!--\; \circ \; -->}$

Let $f\left(x\right)=2\text{sin}\left(3x\right)+4$ for $x\in <!--\; \in \; -->\mathbb{R}$.

Let $g\left(x\right)=5f\left(2x\right)$.

The function $g$ can be written in the form $g\left(x\right)=10\text{sin}\left(bx\right)+c$.

Adam sets out for a hike from his camp at point A. He hikes at an average speed of 4.2 km/h for 45 minutes, on a bearing of 035° from the camp, until he stops for a break at point B.

Adam leaves point B on a bearing of 114° and continues to hike for a distance of 4.6‌ km until he reaches point C.

Adam’s friend Jacob wants to hike directly from the camp to meet Adam at point C .

The quadrilateral ABCD represents a park, where $\text{AB}=120\text{ m}$, $\text{AD}=95\text{ m}$ and $\text{DC}=100\text{ m}$. Angle DAB is 70° and angle DCB is 110°. This information is shown in the following diagram.

A straight path through the park joins the points B and D.

A new path, CE, is to be built such that E is the point on BD closest to C.

The section of the park represented by triangle DCE will be used for a charity race. A track will be marked along the sides of this section.

An archaeological site is to be made accessible for viewing by the public. To do this, archaeologists built two straight paths from point A to point B and from point B to point C as shown in the following diagram. The length of path AB is 185 m, the length of path BC is 250 m, and angle $\text{A}\stackrel{\wedge <!--\; \wedge \; -->}{\text{B}}<!--\; \; -->\text{C}$ is 125°.

The archaeologists plan to build two more straight paths, AD and DC. For the paths to go around the site, angle $\text{B}\stackrel{\wedge <!--\; \wedge \; -->}{\text{A}}<!--\; \; -->\text{D}$ is to be made equal to 85° and angle $\text{B}\stackrel{\wedge <!--\; \wedge \; -->}{\text{C}}<!--\; \; -->\text{D}$ is to be made equal to 70° as shown in the following diagram.

Farmer Brown has built a new barn, on horizontal ground, on his farm. The barn has a cuboid base and a triangular prism roof, as shown in the diagram.

The cuboid has a width of 10 m, a length of 16 m and a height of 5 m.
The roof has two sloping faces and two vertical and identical sides, ADE and GLF.
The face DEFL slopes at an angle of 15° to the horizontal and ED = 7 m .

The roof was built using metal supports. Each support is made from five lengths of metal AE, ED, AD, EM and MN, and the design is shown in the following diagram.

ED = 7 m , AD = 10 m and angle ADE = 15° .
M is the midpoint of AD.
N is the point on ED such that MN is at right angles to ED.

Farmer Brown believes that N is the midpoint of ED.

The following diagram shows a semicircle with centre $\text{O}$ and radius $r$. Points $\text{P, Q}$ and $\text{R}$ lie on the circumference of the circle, such that $\text{PQ}=2r$ and $\text{R}\hat{\text{O}}\text{Q}=\theta$, where $0<\theta <\mathrm{\pi }$.

A farmer owns a plot of land in the shape of a quadrilateral ABCD.

$\text{AB}=105\text{ m, BC}=95\text{ m, CD}=40\text{ m, DA}=70\text{ m}$ and angle $\text{DCB}={90}^{\circ <!--\; \circ \; -->}$.

The farmer wants to divide the land into two equal areas. He builds a fence in a straight line from point B to point P on AD, so that the area of PAB is equal to the area of PBCD.

Calculate

The base of an electric iron can be modelled as a pentagon ABCDE, where:

$$\begin{array}{l}\text{BCDE is a rectangle with sides of length }(x+3)\text{ cm and }(x+5)\text{ cm;}\\ \text{ABE is an isosceles triangle, with AB}=\text{AE and a height of }x\text{ cm;}\\ \text{the area of ABCDE is 222 c}{\text{m}}^{\text{2}}\text{.}\end{array}$$

Insulation tape is wrapped around the perimeter of the base of the iron, ABCDE.

F is the point on AB such that $\text{BF}=\text{8 cm}$. A heating element in the iron runs in a straight line, from C to F.

The following diagram shows a circle with centre $\text{O}$ and radius $3$.

Points $\text{A}$, $\text{P}$ and $\text{B}$ lie on the circumference of the circle.

Chord $\left[\text{AB}\right]$ has length $L$ and $\text{A}\hat{\text{O}}\text{B}=\theta$ radians.

All lengths in this question are in centimetres.

A solid metal ornament is in the shape of a right pyramid, with vertex $\text{V}$ and square base $\text{ABCD}$. The centre of the base is $\text{X}$. Point $\text{V}$ has coordinates $(1, 5, 0)$ and point $\text{A}$ has coordinates $(-1, 1, 6)$.

The volume of the pyramid is $57.2 {\text{cm}}^3$, correct to three significant figures.

A scientist conducted a nine-week experiment on two plants, $A$ and $B$, of the same species. He wanted to determine the effect of using a new plant fertilizer. Plant $A$ was given fertilizer regularly, while Plant $B$ was not.

The scientist found that the height of Plant $A, h_A \text{cm}$, at time $t$ weeks can be modelled by the function $h_A\left(t\right)=\sin (2t+6)+9t+27$, where $0\le t\le 9$.

The scientist found that the height of Plant $B, h_B \text{cm}$, at time $t$ weeks can be modelled by the function $h_B\left(t\right)=8t+32$, where $0\le t\le 9$.

Use the scientist’s models to find the initial height of

At Grande Anse Beach the height of the water in metres is modelled by the function $h(t)=p\cos <!--\; \; -->(q\times <!--\; \times \; -->t)+r$, where $t$ is the number of hours after 21:00 hours on 10 December 2017. The following diagram shows the graph of $h$ , for $0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->72$.

The point $\text{A}(6.25,0.6)$ represents the first low tide and $\text{B}(12.5,1.5)$ represents the next high tide.

A factory packages coconut water in cone-shaped containers with a base radius of 5.2 cm and a height of 13 cm.

The factory designers are currently investigating whether a cone-shaped container can be replaced with a cylinder-shaped container with the same radius and the same total surface area.

A communication tower, T, produces a signal that can reach cellular phones within a radius of 32 km. A straight road passes through the area covered by the tower’s signal.

The following diagram shows a line representing the road and a circle representing the area covered by the tower’s signal. Point R is on the circumference of the circle and points S and R are on the road. Point S is 38 km from the tower and RŜT = 43˚.

The following diagram shows a circle with centre $\text{O}$ and radius $5$ metres.

Points $\text{A}$ and $\text{B}$ lie on the circle and $\text{A}\hat{\text{O}}\text{B}=1.9$ radians.

The following diagram shows a triangle ABC.

$\text{AB}=5\mathrm{c}\mathrm{m},\mathrm{C}\stackrel{^<!--\; ^\; -->}{\mathrm{A}}\mathrm{B}=$ 50° and $\mathrm{A}\stackrel{^<!--\; ^\; -->}{\mathrm{C}}\mathrm{B}=$ 112°

A flat horizontal area, ABC, is such that AB = 100 m , BC = 50 m and angle AĈB = 43.7° as shown in the diagram.

The following diagram shows a circle with centre O and radius 40 cm.

The points A, B and C are on the circumference of the circle and $\mathrm{A}\stackrel{^<!--\; ^\; -->}{\mathrm{O}}\mathrm{C}=1.9\text{ radians}$.

Let $f\left(x\right)=\frac{16}{x}$. The line $L$ is tangent to the graph of $f$ at $x=8$.

$L$ can be expressed in the form r $=\left(\begin{array}{c}8\\ 2\end{array}\right)+t$u.

The direction vector of $y=x$ is $\left(\begin{array}{c}1\\ 1\end{array}\right)$.

A metal sphere has a radius 12.7 cm.

A rocket is travelling in a straight line, with an initial velocity of $140$ m s⁻¹. It accelerates to a new velocity of $500$ m s⁻¹ in two stages.

During the first stage its acceleration, $a$ m s⁻², after $t$ seconds is given by $a\left(t\right)=240\text{sin}\left(2t\right)$, where $0⩽<!--\; ⩽\; -->t⩽<!--\; ⩽\; -->k$.

The first stage continues for $k$ seconds until the velocity of the rocket reaches $375$ m s⁻¹.

The following diagram shows a right-angled triangle, $\text{ABC}$, with $\text{AC}=10\text{cm}$, $\text{AB}=6\text{cm}$ and $\text{BC}=8\text{cm}$.

The points $\text{D}$ and $\text{F}$ lie on $\left[\text{AC}\right]$.
$\left[\text{BD}\right]$ is perpendicular to $\left[\text{AC}\right]$.
$\text{BEF}$ is the arc of a circle, centred at $\text{A}$.
The region $R$ is bounded by $\left[\text{BD}\right]$, $\left[\text{DF}\right]$ and arc $\text{BEF}$.

The following diagram shows a circle, centre O and radius $r$ mm. The circle is divided into five equal sectors.

One sector is OAB, and $\mathrm{A}\stackrel{^<!--\; ^\; -->}{\mathrm{O}}\mathrm{B}=\mathrm{\theta <!--\; \theta \; -->}$.

The area of sector AOB is $20\mathrm{\pi <!--\; \pi \; -->}\text{ m}{\text{m}}^2$.

The following diagram shows a circle with centre $\text{O}$ and radius $1 \text{cm}$. Points $\text{A}$ and $\text{B}$ lie on the circumference of the circle and $\text{A}\hat{\text{O}}\text{B}=2\theta$, where $0<\theta <\frac{\mathrm{\pi }}{2}$.

The tangents to the circle at $\text{A}$ and $\text{B}$ intersect at point $\text{C}$.

The following diagram shows the quadrilateral ABCD.

AB = 6.73 cm, BC = 4.83 cm, BĈD = 78.2° and CD = 3.80 cm.

The following diagram shows part of a circle with centre O and radius 4 cm.

Chord AB has a length of 5 cm and AÔB = θ.